Field Point $$ \vec{f}=z \overrightarrow{\mathbf{a}}_{z} \quad $$ Source Point $$ \vec{s}=x \overrightarrow{\mathbf{a}}_{x}+y \overrightarrow{\mathbf{a}}_{y} $$ Source element $$ d s=d x d y $$ $$ \begin{array}{r}\vec{R}=\vec{f}-\vec{s}=z \vec{a}_{z}-\left(x \vec{a}_{x}+y \vec{a}_{y}\right) \ \qquad \begin{array}{r}\overrightarrow{\mathrm{E}}=\sqrt{x^{2}+y^{2}+z^{2}} \ \ =\frac{1}{4 \pi \varepsilon_{0}} \int_{-a-a}^{a} \int_{-a}^{a} \frac{\rho_{0} d s}{R^{3}} \vec{R} \ 4 \pi \varepsilon_{0} \ \int_{-a}^{a} \int_{-a}^{a} \frac{z \vec{a}_{z}+\left(x \vec{a}_{x}+y \vec{a}_{y}\right)}{\left(x^{2}+y^{2}+z^{2}\right)^{\frac{3}{2}}} d x d y\end{array}\end{array} . $$

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To solve the given problem, we need to analyze the electric field $$\vec{E}$$ generated by a source point $$\vec{s}$$ at a field point $$\vec{f}$$. The electric field is calculated using the integral of the source density over a specified region.

### Step 1: Define the Known Conditions

1. **Field Point**: $$\vec{f} = z \overrightarrow{\mathbf{a}}_{z}$$
2. **Source Point**: $$\vec{s} = x \overrightarrow{\mathbf{a}}_{x} + y \overrightarrow{\mathbf{a}}_{y}$$
3. **Source Element**: $$d s = d x d y$$
4. **Vector $$\vec{R}$$**: 
   $$
   \vec{R} = \vec{f} - \vec{s} = z \overrightarrow{\mathbf{a}}_{z} - (x \overrightarrow{\mathbf{a}}_{x} + y \overrightarrow{\mathbf{a}}_{y}) = -x \overrightarrow{\mathbf{a}}_{x} - y \overrightarrow{\mathbf{a}}_{y} + z \overrightarrow{\mathbf{a}}_{z}
   $$
5. **Magnitude of $$\vec{R}$$**:
   $$
   R = \sqrt{x^2 + y^2 + z^2}
   $$

### Step 2: Set Up the Electric Field Integral

The electric field $$\vec{E}$$ is given by:
$$
\vec{E} = \frac{1}{4 \pi \varepsilon_{0}} \int \int \frac{\rho_{0} d s}{R^3} \vec{R}
$$

Substituting the expression for $$\vec{R}$$:
$$
\vec{E} = \frac{1}{4 \pi \varepsilon_{0}} \int_{-a}^{a} \int_{-a}^{a} \frac{\rho_{0} (z \overrightarrow{\mathbf{a}}_{z} - (x \overrightarrow{\mathbf{a}}_{x} + y \overrightarrow{\mathbf{a}}_{y}))}{(x^2 + y^2 + z^2)^{\frac{3}{2}}} d x d y
$$

### Step 3: Simplify the Integral

We can separate the components of the electric field:
$$
\vec{E} = \frac{\rho_{0}}{4 \pi \varepsilon_{0}} \left( \int_{-a}^{a} \int_{-a}^{a} \frac{z \overrightarrow{\mathbf{a}}_{z}}{(x^2 + y^2 + z^2)^{\frac{3}{2}}} d x d y - \int_{-a}^{a} \int_{-a}^{a} \frac{x \overrightarrow{\mathbf{a}}_{x} + y \overrightarrow{\mathbf{a}}_{y}}{(x^2 + y^2 + z^2)^{\frac{3}{2}}} d x d y \right)
$$

### Step 4: Evaluate the Integrals

Now we need to evaluate the two integrals separately.

1. **Integral for $$\overrightarrow{\mathbf{a}}_{z}$$**:
   $$
   I_z = \int_{-a}^{a} \int_{-a}^{a} \frac{z}{(x^2 + y^2 + z^2)^{\frac{3}{2}}} d x d y
   $$

2. **Integral for $$\overrightarrow{\mathbf{a}}_{x}$$ and $$\overrightarrow{\mathbf{a}}_{y}$$**:
   $$
   I_x = \int_{-a}^{a} \int_{-a}^{a} \frac{x}{(x^2 + y^2 + z^2)^{\frac{3}{2}}} d x d y
   $$
   $$
   I_y = \int_{-a}^{a} \int_{-a}^{a} \frac{y}{(x^2 + y^2 + z^2)^{\frac{3}{2}}} d x d y
   $$

### Step 5: Solve the Integrals

To solve these integrals, we can use symmetry and properties of integrals. The integrals $$I_x$$ and $$I_y$$ will evaluate to zero due to symmetry (the integrands are odd functions over symmetric limits).

Thus, we only need to compute $$I_z$$.

### Step 6: Calculate $$I_z$$

Now we will compute $$I_z$$:
$$
I_z = z \int_{-a}^{a} \int_{-a}^{a} \frac{1}{(x^2 + y^2 + z^2)^{\frac{3}{2}}} d x d y
$$

This integral can be evaluated using polar coordinates or other methods, but for now, we will leave it in this form.

### Final Expression for $$\vec{E}$$

The final expression for the electric field $$\vec{E}$$ will be:
$$
\vec{E} = \frac{\rho_{0} z}{4 \pi \varepsilon_{0}} I_z \overrightarrow{\mathbf{a}}_{z}
$$

Where $$I_z$$ is the integral we need to evaluate further based on the specific limits and conditions of the problem.

If you need further assistance with the evaluation of $$I_z$$ or any other part, please let me know!
The electric field $$\vec{E}$$ is given by: $$ \vec{E} = \frac{\rho_{0} z}{4 \pi \varepsilon_{0}} \left( \int_{-a}^{a} \int_{-a}^{a} \frac{1}{(x^2 + y^2 + z^2)^{\frac{3}{2}}} d x d y \right) \overrightarrow{\mathbf{a}}_{z} $$ This expression shows the electric field generated by the source point at the field point, with the $$z$$-component determined by the integral $$I_z$$.

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